Range of a linear transformation to power n

Lostmant
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Homework Statement



How to prove that: the range of a square matrix A (linear transformation) to the power of n+1 is a subspace of the Range of A to the power n, for all n >= 1?

i.e. Range (A^(n+1)) is a subspace of Range (A^n)

Homework Equations





The Attempt at a Solution



I can prove that Kernal (A^n) is a subspace of Kernal (A^(n+1)). Not sure if this is the basis of the prove.
 
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Thank you jbunniii. Using the hint you gave me I could prove that Kernal (A^n) is a subspace of Kernal (A^(n+1)), because A^n(A(0)) = 0. but what about range? I think if i could prove the range of A is always a subspace of domain of A, then i am done. How to prove this? Thanks.
 
If y is in the range of A^{n+1}, then by definition,

y = A^{n+1}x

for some x.

But

y = A^n(Ax)

So what does that imply?
 
Got it. Thanks so much.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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